3.1 \(\int \cos ^3(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=125 \[ -\frac{a (5 A+4 B) \sin ^3(c+d x)}{15 d}+\frac{a (5 A+4 B) \sin (c+d x)}{5 d}+\frac{a (A+B) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 a (A+B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3}{8} a x (A+B)+\frac{a B \sin (c+d x) \cos ^4(c+d x)}{5 d} \]

[Out]

(3*a*(A + B)*x)/8 + (a*(5*A + 4*B)*Sin[c + d*x])/(5*d) + (3*a*(A + B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*(A
+ B)*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (a*B*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) - (a*(5*A + 4*B)*Sin[c + d*
x]^3)/(15*d)

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Rubi [A]  time = 0.167091, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {2968, 3023, 2748, 2633, 2635, 8} \[ -\frac{a (5 A+4 B) \sin ^3(c+d x)}{15 d}+\frac{a (5 A+4 B) \sin (c+d x)}{5 d}+\frac{a (A+B) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 a (A+B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3}{8} a x (A+B)+\frac{a B \sin (c+d x) \cos ^4(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + a*Cos[c + d*x])*(A + B*Cos[c + d*x]),x]

[Out]

(3*a*(A + B)*x)/8 + (a*(5*A + 4*B)*Sin[c + d*x])/(5*d) + (3*a*(A + B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*(A
+ B)*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (a*B*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) - (a*(5*A + 4*B)*Sin[c + d*
x]^3)/(15*d)

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \cos ^3(c+d x) (a+a \cos (c+d x)) (A+B \cos (c+d x)) \, dx &=\int \cos ^3(c+d x) \left (a A+(a A+a B) \cos (c+d x)+a B \cos ^2(c+d x)\right ) \, dx\\ &=\frac{a B \cos ^4(c+d x) \sin (c+d x)}{5 d}+\frac{1}{5} \int \cos ^3(c+d x) (a (5 A+4 B)+5 a (A+B) \cos (c+d x)) \, dx\\ &=\frac{a B \cos ^4(c+d x) \sin (c+d x)}{5 d}+(a (A+B)) \int \cos ^4(c+d x) \, dx+\frac{1}{5} (a (5 A+4 B)) \int \cos ^3(c+d x) \, dx\\ &=\frac{a (A+B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{a B \cos ^4(c+d x) \sin (c+d x)}{5 d}+\frac{1}{4} (3 a (A+B)) \int \cos ^2(c+d x) \, dx-\frac{(a (5 A+4 B)) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac{a (5 A+4 B) \sin (c+d x)}{5 d}+\frac{3 a (A+B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a (A+B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{a B \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac{a (5 A+4 B) \sin ^3(c+d x)}{15 d}+\frac{1}{8} (3 a (A+B)) \int 1 \, dx\\ &=\frac{3}{8} a (A+B) x+\frac{a (5 A+4 B) \sin (c+d x)}{5 d}+\frac{3 a (A+B) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a (A+B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{a B \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac{a (5 A+4 B) \sin ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.24435, size = 77, normalized size = 0.62 \[ \frac{a \left (-160 (A+2 B) \sin ^3(c+d x)+480 (A+B) \sin (c+d x)+15 (A+B) (12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x)))+96 B \sin ^5(c+d x)\right )}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + a*Cos[c + d*x])*(A + B*Cos[c + d*x]),x]

[Out]

(a*(480*(A + B)*Sin[c + d*x] - 160*(A + 2*B)*Sin[c + d*x]^3 + 96*B*Sin[c + d*x]^5 + 15*(A + B)*(12*(c + d*x) +
8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)])))/(480*d)

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Maple [A]  time = 0.05, size = 128, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({\frac{aB\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+aA \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +aB \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +{\frac{aA \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+cos(d*x+c)*a)*(A+B*cos(d*x+c)),x)

[Out]

1/d*(1/5*a*B*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+a*A*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+
3/8*d*x+3/8*c)+a*B*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+1/3*a*A*(2+cos(d*x+c)^2)*sin(d
*x+c))

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Maxima [A]  time = 1.0119, size = 167, normalized size = 1.34 \begin{align*} -\frac{160 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a - 15 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 32 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a - 15 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*cos(d*x+c))*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

-1/480*(160*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))
*A*a - 32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*B*a - 15*(12*d*x + 12*c + sin(4*d*x + 4*c)
+ 8*sin(2*d*x + 2*c))*B*a)/d

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Fricas [A]  time = 1.76045, size = 239, normalized size = 1.91 \begin{align*} \frac{45 \,{\left (A + B\right )} a d x +{\left (24 \, B a \cos \left (d x + c\right )^{4} + 30 \,{\left (A + B\right )} a \cos \left (d x + c\right )^{3} + 8 \,{\left (5 \, A + 4 \, B\right )} a \cos \left (d x + c\right )^{2} + 45 \,{\left (A + B\right )} a \cos \left (d x + c\right ) + 16 \,{\left (5 \, A + 4 \, B\right )} a\right )} \sin \left (d x + c\right )}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*cos(d*x+c))*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

1/120*(45*(A + B)*a*d*x + (24*B*a*cos(d*x + c)^4 + 30*(A + B)*a*cos(d*x + c)^3 + 8*(5*A + 4*B)*a*cos(d*x + c)^
2 + 45*(A + B)*a*cos(d*x + c) + 16*(5*A + 4*B)*a)*sin(d*x + c))/d

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Sympy [A]  time = 3.58734, size = 333, normalized size = 2.66 \begin{align*} \begin{cases} \frac{3 A a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 A a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 A a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{3 A a \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{2 A a \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac{5 A a \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac{A a \sin{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac{3 B a x \sin ^{4}{\left (c + d x \right )}}{8} + \frac{3 B a x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac{3 B a x \cos ^{4}{\left (c + d x \right )}}{8} + \frac{8 B a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac{4 B a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac{3 B a \sin ^{3}{\left (c + d x \right )} \cos{\left (c + d x \right )}}{8 d} + \frac{B a \sin{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac{5 B a \sin{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text{for}\: d \neq 0 \\x \left (A + B \cos{\left (c \right )}\right ) \left (a \cos{\left (c \right )} + a\right ) \cos ^{3}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+a*cos(d*x+c))*(A+B*cos(d*x+c)),x)

[Out]

Piecewise((3*A*a*x*sin(c + d*x)**4/8 + 3*A*a*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*A*a*x*cos(c + d*x)**4/8 +
3*A*a*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 2*A*a*sin(c + d*x)**3/(3*d) + 5*A*a*sin(c + d*x)*cos(c + d*x)**3/(
8*d) + A*a*sin(c + d*x)*cos(c + d*x)**2/d + 3*B*a*x*sin(c + d*x)**4/8 + 3*B*a*x*sin(c + d*x)**2*cos(c + d*x)**
2/4 + 3*B*a*x*cos(c + d*x)**4/8 + 8*B*a*sin(c + d*x)**5/(15*d) + 4*B*a*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) +
3*B*a*sin(c + d*x)**3*cos(c + d*x)/(8*d) + B*a*sin(c + d*x)*cos(c + d*x)**4/d + 5*B*a*sin(c + d*x)*cos(c + d*
x)**3/(8*d), Ne(d, 0)), (x*(A + B*cos(c))*(a*cos(c) + a)*cos(c)**3, True))

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Giac [A]  time = 1.10095, size = 151, normalized size = 1.21 \begin{align*} \frac{3}{8} \,{\left (A a + B a\right )} x + \frac{B a \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac{{\left (A a + B a\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac{{\left (4 \, A a + 5 \, B a\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac{{\left (A a + B a\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{{\left (6 \, A a + 5 \, B a\right )} \sin \left (d x + c\right )}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*cos(d*x+c))*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

3/8*(A*a + B*a)*x + 1/80*B*a*sin(5*d*x + 5*c)/d + 1/32*(A*a + B*a)*sin(4*d*x + 4*c)/d + 1/48*(4*A*a + 5*B*a)*s
in(3*d*x + 3*c)/d + 1/4*(A*a + B*a)*sin(2*d*x + 2*c)/d + 1/8*(6*A*a + 5*B*a)*sin(d*x + c)/d